Universal Totally Ordered Sets

Luke Adams

May 11, 2018

Abstract

In 1874, Georg Cantor published an article in which he proved thatthe set of algebraic numbers are countable, and the set of real numbersare uncountable. This, at the time controversial article, marked thebeginning of modern set theory, and it finally gave mathematiciansthe notation to explore the infinite. In the intervening century anda half, set theory has blossomed into a central part of mathematics,often acting as a language with which to formalize other branches ofmathematics [5].

In this paper, we explore some basic concepts in set theory, andthen consider a result that was proven in the Introduction to HigherMathematics notes: every countable, totally ordered set is embeddableinto the rational numbers. We generalize this idea to larger cardinalitiesand introduce the notion of ℵα-universal sets. We show that ηα-sets are,in fact, ℵα-universal, and conclude by providing a construction of theminimal ηα-set whenever ℵα is a regular ordinal, and the generalizedcontinuum hypothesis holds.

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Contents

1 Introduction 31.1 Ordered Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Cardinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Ordinal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Order Types . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Maximal POsets . . . . . . . . . . . . . . . . . . . . . . . . . 101.6 Distinct Order Types . . . . . . . . . . . . . . . . . . . . . . . 11

2 Countable Totally Ordered Sets 13

3 Cofinality 16

4 ℵα-universal sets 184.1 ηα-sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Constructing a minimal ηα-set . . . . . . . . . . . . . . . . . 21

5 Conclusion 24

References 25

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1 Introduction

The study of set theory is due primarily to Georg Cantor in 1874. In thisyear, Cantor published a paper that proved the existence of a one-to-onecorrespondence between the set of algebraic numbers and the set of naturalnumbers. In the same paper, he also showed that no such correspondenceexists between the set of natural numbers and the set of real numbers. A fewyears later, Cantor would coin the terms ‘countable’ and ‘uncountable’ todescribe these relationships. Cantor went on to spearhead the developmentof set theory, creating an entirely new field of study in his wake [5].

Nearly 40 years later, in 1914, Felix Hausdorff introduced the notion ofuniversally ordered sets, which contain copies of all totally ordered sets of aparticular cardinality. Hausdorff showed that such universally ordered setsare guaranteed to exist. Further, Hausdorff provided a way to constructthese sets in a select number of cases [2].

In this paper we consider the properties of infinite ordered sets. We beginby reviewing some basic notions in set theory, and prove several nonintuitivefacts about infinite sets. We then define universally ordered sets, and discusstheir implications. We conclude by examining a construction of universallyordered sets that is closely related to the ones that Hausdorff published.

1.1 Ordered Sets

The key objects in all of our explorations will be ordered sets: a set S andan order ≤. We already have many preconceived notions about how an ordershould behave. For example, we expect that an order should be transitivein the same way that an equivalence relation is. That is, if a ≤ b and b ≤ c,then we expect that a ≤ c. We use these expectations to guide of definitionof an order.

Definition 1 ([4]). Let S be a set, and let ≤ be a relation on S. Furtherlet a, b, c ∈ S be arbitrary elements. We say that ≤ is a partial order onS if it satisfies three properties:

1. Reflexivity : a ≤ a.

2. Transitivity : If a ≤ b and b ≤ c, then a ≤ c.

3. Antisymmetry. If a ≤ b and b ≤ a, then a = b.

We call the pair (S,≤) a partially ordered set, or a POset. We say that

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two elements a, b ∈ S are comparable if either a ≤ b or b ≤ a. Otherwise,we say that a and b are incomparable, and we write a||b.

The relation ≤ is called a total order if every pair of elements in Sis comparable. If this extra condition holds, we call the pair (S,≤) atotally ordered set.

Let S be a totally ordered set, and let A,B ⊆ S be subsets. We willwrite A < B when a < b for all elements a ∈ A and b ∈ B. We say that Aand B are neighboring if there is no element s ∈ S such that A < s < B.

Suppose s ∈ S is an element. Then we say that s is a successor elementif there is an element t ∈ S such that t < s, and t and s are neighboring. Ifthere is no such element, we say that s is a lower-limit element. Dually, wesay that s is a predecessor element if there is a t such that s < t, and s andt are neighboring. If no such t exists, then s is an upper-limit element.

We say that a ∈ A is the least element of the subset A if a ≤ a′ for alla′ ∈ A. There is also the dual notion of the greatest element of a subset. Ifevery subset of S has a least element, then the structure of S is very limited.These sets are very important, so we given them a special name:

Definition 2 ([3]). Suppose S is a totally ordered set with order ≤.Then we say that ≤ is a well-order and (S,≤) is a well-ordered set ifevery subset of S has a least element. Analogously, we say that ≤ is areverse well-order and (S,≤) is a reverse well-ordered set if every subsetof S has a greatest element.

The structure of a well-ordered set means that all of the elements of Scan be listed in order starting with the least element: S = {s0, s1, s2, . . .}.

1.1.1 Examples of Ordered Sets

Before continuing, we consider several examples of ordered sets. We encounterordered sets in all aspects of math, sometimes without realizing. Otherexamples are more contrived, but they highlight important aspects of orderedset theory. This library of specific examples will allow us to ground abstractarguments in particular examples.

The set of natural numbers, N, under the standard ordering is well-ordered1, however the integers, Z, are only totally ordered because somesubsets (including Z itself) do not have least elements. The negative integers,

1Actually, we must assume that the natural numbers are well-ordered and introducea new axiom to this effect. But we almost always make this assumption, because it isequivalent to assuming that standard induction is valid [3].

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{1,2,3}

{1,3}{1,2} {2,3}

{2}

∅

{1} {3}

Figure 1: Subset lattice for the set {1, 2, 3}. The lines between sets indicatea subset relationship, with the set that is higher containing the set that islower.

Z−, are reverse well-ordered. Both the rational numbers, Q, and the realnumbers, R, are totally ordered with the standard ordering.

The operations of set theory can also naturally create orders. The powerset of a set S is the collection of all subsets of S. We write this as P(S).Consider the set S = {1, 2, 3}, and observe that

P(S) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}.

The power set of S is a partial order under set inclusion. That is, if A,B ⊆ S,then A ≤ B if and only if A ⊆ B. When the number of elements in a partialset is finite, we can create diagrams to indicate the order structure of theset. Such a diagram for the set S is shown in Figure 1.

The alphabetic ordering of words forms a total order. To determinewhich word should come first in an alphabetical listing, we compare the firstletter of each word, and the word with an earlier first letter is listed first. Ifthe first letters are the same, we compare the second letters, and so on. Forexample, consider the two words “love” and “long.” The first two letters ofeach word are the same, so we compare the third letter of each word, andbecause ‘n’ < ‘v’, we conclude that “long” < “love.”

Notice that in the alphabetical ordering, we are making use of theunderlying order of the letters, and then viewing a word as simply an orderedtuple of letters. This style of ordering is called lexicographical ordering, andwe can also use it to order other sets consisting of ordered tuples. For example,consider the set N× N, ordered lexicographically. This set is well-ordered

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because every subset has a least element, and thus we can list the elementsin order:

(1, 1) < (1, 2) < · · · < (2, 1) < (2, 2) < · · · < (3, 1) < (3, 2) < · · · .

Notice that the elements (1, 1), (2, 1), . . . are lower-limit elements, whileevery other element is a successor. Every element in N× N is a predecessorelement.

1.1.2 Isomorphisms and Embeddings

Suppose A and B are POsets with orders ≤A and ≤B respectively. Thenwe say that A and B are order-isomorphic and we write A ' B if there is abijection f : A→ B with the property that, for every x, y ∈ A,

x ≤A y =⇒ f(x) ≤B f(y).

If two sets are order-isomorphic, then they are identical from a set-theoretic point of view. This means that anything we prove about a set A isalso true for every set B that is order-isomorphic to A.

Now suppose that, instead of a bijection, the function f : A→ B is anorder preserving injection. Then we say that A is embeddable into B, andwe write A � B. In other words, A is embeddable into B exactly when A isorder-isomorphic to some subset of B.

The notation that we use for embeddablity is suggests that that thisrelation might, in fact, form an order. To determine if this is true, we mustcheck if the � satisfies the three conditions of a partial order: reflexivity,transitivity, and anti-symmetry.

Theorem 3. The relation � is reflexive and transitive.

Proof. Let A, B, and C be arbitrary POsets. Note that the identity functionIA(a) = a is clearly an embedding of A into A, and thus A � A.

Now suppose that A � B and B � C. Then it follows that there areembeddings f : A → B and g : B → C. We claim that the composition(g ◦ f) : A→ C is also an embedding, and thus A � C.

To see why this is true, suppose a, a′ ∈ A with a ≤ a′. Then there areimages b = f(a) and b′ = f(a′) with the property that b < b′. Then finally,we have that g(b) ≤ g(b′), and thus (g ◦ f)(a) ≤ (g ◦ f)(a′), as required.

Thus, if � was anti-symmetric, it would form a partial order on orderedsets. However, this is not true, and it then follows that � is not a partial

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order. We can see the lack of anti-symmetry by considering the two setsA = [0, 1] and B = (0, 1). Clearly these sets have different order properties(particularly at the endpoints), so they are not order-isomorphic.

However, if we consider the embedding functions f : A→ B and g : B →A defined by

f(x) =1

4x+

1

4and g(x) = x, (1)

we can easily verify that A � B and B � A. Therefore, � is not anti-symmetric, and thus not a partial order. We refer to a relation that isreflexive and transitive, but not anti-symmetric as a quasi-order.

1.2 Cardinal Numbers

Suppose we would like to determine if two sets A and B have the samenumber of elements. If both A and B are finite, then we can simply countthe number of elements in each set. However, this notion breaks down whenthe sets become infinite. This was Cantor’s dilemma.

To resolve this difficulty, we say that two sets A and B have the samecardinality if there is an isomorphism f : A → B. In this case, we write|A| = |B|. For example, in Math 260, we showed that |N| = |Q| and that|N| < |R|.

But what is |A|? If the set A is finite, then |A| is the number of elementsin the set, so |A| ∈W. However, the answer is less obvious if A is infinite.Clearly the “size” of A cannot be represented by an integer, so |A| must befrom an even larger class of numbers that is somehow expanded to includeinfinite numbers.

We call these cardinal numbers. The class of cardinal numbers containsall W, but it also has an infinite number of other elements, each representingan infinite size. The least infinite cardinal is |N| which we typically denoteas ℵ0 or just ℵ. The next larger cardinal numbers are ℵ1, ℵ2, and so on.

The cardinal numbers are well-ordered , so we can list them in ascendingorder:

0 < 1 < 2 < · · · < ℵ0 < ℵ1 < ℵ2 < · · · .

1.2.1 Cardinal Arithmetic

We can use the cardinal numbers to do arithmetic. In fact, in very formaltreatments of mathematics, all numbers are actually sets. Before we give thedefinitions of the arithmetic operations on cardinal numbers, we first needto define some notation.

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Let X and Y be two sets. We define the disjoint union of X and Y as

X∪Y = ({0} ×X) ∪ ({1} × Y ).

Essentially, this disjoint union ensures that, if X and Y have some elementsin common, then the common elements are “repeated twice” in the union.We also define XY to be the collection of all functions from X into Y . Withthese definitions, we are now ready to define our cardinal arithmetic operators[3].

Definition 4. Suppose X and Y are sets, then we will define addition,multiplication, and exponentiation of cardinal numbers as

|X|+ |Y | = |X∪Y |, |X| · |Y | = |X × Y |, and |X||Y | =∣∣XY

∣∣ .There are a few interesting consequences of these definitions. For example,

suppose X is a set. Then there is a natural bijection between P(X) and 2X ,where 2 = {0, 1}. In particular, suppose A ∈ P(X). Then we pair A withthe function f : X → 2 defined by

fA(x) =

{1, x ∈ A;

0, x 6∈ A.

This means that |P(X)| = 2|X|. As we will see in next section, this relation-ship can be very useful in some circumstances.

1.2.2 The Continuum Hypothesis

Let c = |R| denote the cardinality of the real numbers. There is no bijectionbetween the natural numbers and the real numbers, so we know that ℵ0 < c.However, we can construct a bijection between the family of subsets of thenatural numbers and the real numbers, and thus we have |P(N)| = 2ℵ0 = c.

But neither of these facts actually tell us the cardinality of the realnumbers. We could have c = ℵ1, but we could also have c = ℵ5.

It turns that this problem is undecidable. That is, given the standardaxioms of set theory, it is impossible to prove the cardinality of the realnumbers. Thus, we must make some arbitrary choice, and introduce anew axiom to enforce this choice. One such choice we could make is theContinuum Hypothesis [1].

Axiom 5. Continuum Hypothesis (CH) The cardinality of the real numbersis c = ℵ1.

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There is a order-isomorphism between P(N) and R, so this means that,assuming the CH, P(N) = 2ℵ0 = ℵ1. That is, the power set of a countableset is only “one infinity bigger” than the set itself!

Notice that the CH still has not told us anything about the relative sizesof larger ℵ sets. However, we can use the notion of power sets to createa more general axiom that specifies the relationships between all infinitecardinal numbers:

Axiom 6. Generalized Continuum Hypothesis (GCH) Suppose S is a setwith |S| = ℵα. Then P(S) = 2ℵα = ℵα+1.

In what follows, we will not, in general assume that the CH or the GCHholds. However, there will be several circumstances where we will find itadvantageous to assume these axioms to allow us to push our argumentsfurther. In every case where we rely on the CH or GCH, we will make anexplicit comment about their use.

1.3 Ordinal Numbers

Cardinality allows us to compare the sizes of arbitrary sets, however we cando even better if the sets are well-ordered. If two well-ordered sets A and Bare order isomorphic, we say that A and B have the same ordinality, and wewrite ||A|| = ||B||.

For a finite set A, there is exactly one well-ordered set (up to order-isomorphism), and thus ||A|| = |A|. That is, the cardinality of A is equivalentto the ordinality of A. However, this is not true when A is infinite. In fact,for a given cardinality, there are an infinite number of well-ordered sets withdifferent ordinalities.

We call these objects ordinal numbers, and we typically denote themusing lowercase Greek letters (e.g. α, β, ν, µ). In fact, we define the cardinalnumbers to be a subset of the ordinal numbers such that ℵα = ωα [3].

The ordinal numbers are themselves well-ordered, so we can list theelements, and also indicate the cardinality of each number:

0 < 1 < 2 < · · · < ω0 < ω0 + 1 < ω0 + 2 < · · ·︸ ︷︷ ︸ℵ0

< ω1 < ω1 + 1 < ω1 + 2 < · · ·︸ ︷︷ ︸ℵ1

< · · ·

The ordinal numbers are in fact sets, and their construction gives themthe somewhat unusual property that, for all ordinals µ and ν,

µ < ν ⇐⇒ µ ∈ ν.

For a much more detailed discussion of the construction of the ordinal(and cardinal) numbers, see [3].

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1.4 Order Types

When we defined the ordinal numbers, we only considered order-isomorphismsbetween well-ordered sets. However, we can relax this restriction to talkabout the structure of all ordered sets. As we noted earlier, if two sets areorder-isomorphic, then they are identical from a set-theoretic perspective.With this in mind, we will organize POsets into order types such that everyset within a given order type is order-isomorphic to every other set in theorder type.

Definition 7. Suppose A and B are POsets with A ' B. Then wesay that A and B share an order type, and we write tp(A) = tp(B). Iftp(A) = α, then we say that A is a realization of α.

We refer to some order types so much that we assign them special names.A few are

tp(N) = ω, tp(Q) = η, and tp(R) = λ. (2)

Suppose α and β are order types, and A and B are realizations of α andβ respectively. We say that α is embeddable in β, and we write α � β if A isembeddable into B [1].

1.5 Maximal POsets

Suppose S is some set with a partial order ≤, and consider the set X≤ ∈P(S × S) defined by

X≤ = {(s, t) | s, t ∈ S and s ≤ t}. (3)

That is, X≤ is the set of all pairs of elements in S for which the first element isless than the second element. The set X≤ is called the graph of ≤. In formaltreatments of ordered set theory, the graph of a relation is the definition ofthe relation.

Let’s consider the collection X ⊆ P(S × S) of all possible partial ordersof S. This family of sets is partially ordered under inclusion.

Theorem 8. The collection X has a least element. Namely, the identityrelation

IS = {(s, s) | s ∈ S}. (4)

Proof. Suppose X ∈ X is some partial order of S. Then X must be reflexive,and thus (s, s) ∈ X for all s ∈ S. It then follows that IS ⊆ X, and thus ISis the least element of X .

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We have shown that X has a least element, and now we would like toinvestigate what it means for an order to be maximal in X . We claim thatan order is maximal in X if and only if it is a total order. Intuitively, thisshould make some sense: in a total order, every pair of elements has a definedorder, so there are no more pairs of elements that could be added to thegraph of ≤ to create a larger graph. Let’s formalize this intuition.

Theorem 9. Suppose S is a set and ≤ is a partial order of S with the graphX≤. Then ≤ is a total order exactly when X≤ is maximal in X .

Proof. First suppose that ≤ is a total order; we need to show that its graphis maximal in X . To that end, suppose ≤′ is another partial order such thatX≤ ⊆ X≤′ . We will show that this is actually an equality.

Let (s, t) ∈ X≤′ be a pair of arbitrary, distinct elements. If (s, t) ∈ X≤,we are done, so suppose this is not the case. Then s � t, and thus we musthave t ≤ s. However, this implies that t ≤′ s, and thus s = t, contradictingour assumption that s and t are distinct. We therefore conclude that ≤ ismaximal in X .

Now suppose that ≤ is not a total order; we would like to conclude thatX≤ is not maximal in X either.

Let � be some well-ordering of S. Then we can use this ordering toconstruct a lexicographic total ordering �L on the set 2S .

Now if s ∈ S, we define s ∈ 2S as

s(t) =

{0, a � x;

1, a ≤ x.

Notice that s is just the indicator function for the subset s = {t ∈ S : t <s} ⊆ S. It follows that x 7→ x is injective (i.e. if x and y map to the sameplace, then x = y). Let A ⊆ 2A be the image of A under this mapping.

It is easy to see that, for all x, y ∈ A, we have

x ≤ y =⇒ x ⊆ y =⇒ x �L y.

Therefore, we can extend ≤ with a new order ≤′ by letting x ≤′ y meanx �L y. Since �L is a total order on 2A, we know that ≤′ is a total order onA. It follows that X≤′ is maximal in X , and thus X≤ is not maximal.

1.6 Distinct Order Types

We know that there are an infinite number of POsets of a particular cardinality.This is because, given a particular POset, we can replace some of the symbols

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in the POset to produce a new POset. However, these two sets would sharethe same order type, and thus they would be, for all intents and purpose,identical. In this section, our goal will be to determine how many uniqueorder types exist for a particular cardinality.

Let T be a totally ordered set, and let t ∈ T be an element. Then theinitial segment defined by t is the set of all s ∈ T such that s < t. Similarly,the final segment defined by t is the set of all s ∈ T for which t < s. We write(−∞, t) and (t,∞) to indicate the initial and final segments of t respectively.

We will let T [I] be the set of elements in T whose initial segments do nothave a greatest element, and we let T [F ] be the set of elements in T whosefinal segments do not have a least element. Notice that if T and S are twototally ordered sets with T ' S, then T [I] = S[I] and T [F ] = S[F ].

Consider the set A = ℵα ×W. The structure of A is

(0, 0) < (0, 1) < · · · < (1, 0) < (1, 1) < · · · < (2, 0) < (2, 1) < · · · .

Notice that every element (ν, n) ∈ A has a well defined successor; namely(ν, n+1). Thus every final segment in A has a least element, and so A[F ] = ∅.However, this is not the case for all initial segments. In particular, (ν, 0) doesnot have a predecessor, and thus A[I] = ℵα × {0}. You might be concernedabout the element (0, 0) ∈ A[I], but notice that (−∞, (0, 0)) = ∅, and thusthe initial segment does not have a greatest element (i.e. has no elements atall!).

Now we have the notation to answer the key question of this section:how many non-order-isomorphic totally ordered sets with cardinality ℵα?Consider X = P(ℵα) = P(ωα), the family of all subsets of ℵα. Our goalis to construct, for every X ∈ X , a totally ordered set TX with cardinality|TX | = ℵα, and with the property that TX = TY if and only if X = Y . Thatis, each TX should define a different order type of cardinality ℵα.

For each X ∈ X , we define TX ⊆ ℵα ×Q such that

TX = (ℵα ×W) ∪(X ×

{1

2,1

4, . . .

}).

Essentially, every TX is the set A that we considered previously, along withsome extra elements that limit down to (x, 0) for every x ∈ X. For example,suppose X = {1}. Then the ordering of TX is

(0, 0) < (0, 1) < · · · < (1, 0) < · · · < (1, 14) < (1, 12) < (1, 1) < · · · < (2, 0) < · · · .

Notice that |Q| = ℵ0, and thus |ℵα ×Q| = ℵα. It then clearly follows that|TX | ≤ ℵα.

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Observe that TX [I] = ℵα × {0} just as for our set A, however, unlikebefore, TX [F ] is nonempty because the elements (x, 0) for x ∈ X do not havea successor. It follows that TX [F ] = X × {0}.

Now suppose X,Y ∈ X are two sets with the property that an orderisomorphism f : TX → TY exists. We would like to conclude that X = Y ,and thus none of the TX sets are order-isomorphic to each other.

Notice that, by the definition of our set T [I], we must have that

f(ℵα × {0}) = ℵα × {0},

but this set is well-ordered, so f must be the identity function for this subsetof the elements of TX and TY . It then follows that

f(X) = Y,

but we have already concluded that this subset of f is the identity function,so we have X = Y , as required.

We have shown that there are at least P(ℵα) = 2ℵα non-order-isomorphictotally ordered sets with cardinality ℵα. Have we found them all, or arethere more order types out there? The next theorem tells us that we haveindeed found all the possible sets that match our criteria.

Theorem 10. For any infinite cardinal ℵα, there are exactly 2ℵα order typesof totally ordered sets with cardinality ℵα.

Proof. Suppose S is a set with |S| = ℵα, and suppose that ≤1 and ≤2

are any two total orders of S. Then (S,≤1) and (S,≤2) will be order-isomorphic if and only if ≤1 and ≤2 have the same graph. However, recallthat |S × S| = |S| = ℵα by our cardinal arithmetic rules, and thus there canbe no more than 2ℵα subsets.

We have just proven that there are at most 2ℵα non-order-isomorphictotally ordered set with cardinality ℵα. Previously, we saw that there are atleast this many sets, so we conclude that there are exactly 2ℵα such sets.

If we assume the GCH, then we can say even more.

Corollary 11. Suppose the GCH holds. Then there are exactly ℵα+1 distinctorder types of cardinality ℵα.

2 Countable Totally Ordered Sets

In the final section of the Introduction to Higher Mathematics notes, weproved that every countable, totally ordered set is order-isomorphic to a

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subset of the rational numbers [4]. This means that, in some sense, theset of rational numbers Q contain every possible countably infinite, totallyordered set. We would like to extend this notion of prototypical sets to largercardinalities, and use these sets to study the structure of totally ordered sets.

To aid in this generalization, we introduce the notion of universallyordered sets [1].

Definition 12. Suppose S is a totally ordered set. Then S is ℵα-universal if every totally ordered set A with |A| ≤ ℵα is order-isomorphicto a subset of S. Equivalently, A must be embeddable into S.

In this new language, we rephrase our claim: we proved in Introduction toHigher Mathematics that Q is ℵα-universal.

Notice that the every countable, totally ordered set is embeddable in thereal numbers as well. That is, R is also ℵ0 universal. But this is a not asnice a result as the first one because the cardinality of R is greater than ℵ0.In other words, R has more elements than needed for this property to hold.With this in mind, we say that a set S is a minimal ℵα-universal set if S isℵα-universal, and |S| = ℵα.

Understanding the details of this argument are critical for what follows,so, we state the Introduction to Higher Mathematics claim, and review theproof given in the notes [4].

Theorem 13. Every countable, totally ordered set is order-isomorphic to asubset of the rational numbers.

Equivalently, the set of rational numbers is ℵ0-universal.

Proof. Suppose the set A = {a1, a2, a3, . . .} is a countable, totally orderedset. We will inductively construct an embedding function f : A→ Q.

We assign f(a1) be any rational number, say zero. Now suppose that wehave defined f(a1), f(a2), . . . , f(an) in such a way that the order relations ofA and Q have been preserved. We need to define f(an+1) so that f continuesto preserve order.

To do this, we partition the set {a1, . . . , an} into two subsets:

A+n+1 = {ai : i ≤ n and ai > an+1}

A−n+1 = {ai : i ≤ n and ai < an+1}.

In Q, we clearly have that f(A−n+1) < f(A+n+1), and so we can choose a

number q ∈ Q such that f(A−n+1) < {q} < f(A+n+1). We then let f(an+1) = q,

which preserves the order relations of A and Q. When carried out infinitely,

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the resulting function will be defined for all of A, and it will be an order-preserving injective function.

Therefore, we have successfully embedded A into Q, and thus Q is ℵα-universal.

We know that Q is ℵ0-universal, but what are the unique properties thatmake it ℵ0-universal? One property that was extremely important in theprevious proof was that, given any two rational numbers a and b with a < b,there is always a third rational number c such that a < c < b.

Theorem 14. Suppose T is a totally ordered set. The T will be orderisomorphic to Q if and only if the following hold:

1. |T | = |Q|;

2. For all pairs of finite subsets A ⊆ T and B ⊆ T , if A < B, then thereis a t ∈ T such that A < {t} < B.

Proof. First suppose that f : T → Q is an order isomorphism. Then clearly|T | = |Q|. To verify the second condition, suppose that A ⊆ T and B ⊆ Tare finite subsets of T with the property A < B. Observe that f(A) < f(B),and thus there is an element q ∈ Q such that f(A) < {q} < f(B). Thenthere is some element t ∈ T such that f(t) = q, and clearly A < {t} < B.

Now suppose that the two conditions hold. We will verify that T isorder isomorphic to Q by inductively constructing an order isomorphismg : T → Q. Both T and Q are countable, so we can enumerate the elements:T = {t1, t2, . . .} and Q = {q1, q2, . . .}. We pair the elements t1 and q1, sothat g(t1) = q1. Now we repeat the following algorithm:

1. Consider the first element tn+1 ∈ T that has not yet been assigned animage. Form the sets

T+n+1 = {ti | i ≤ n and ti > tn+1}T−n+1 = {ti | i ≤ n and ti < tn+1}.

In Q, it is clear that g(T−n+1) < g(T+n+1), and so we choose a q ∈ Q

such that g(T−n+1) < {q} < g(T+n+1), and assign g(tn+1) = q. This

assignment is consistant with the orders of both T and Q.

2. Now consider the first element qj ∈ Q that has not been assigned animage. Form the sets

Q+n+1 = {qi | i ≤ n and qi > qn+1}

Q−n+1 = {qi | i ≤ n and qi < qn+1}.

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In T , it is clear that g−1(Q−n+1) < g−1(Q+n+1), and so we choose a t ∈ T

such that g−1(Q−n+1) < {t} < g−1(Q+n+1), and assign g−1(tn+1) = q.

Once again, this assignment is consistant with the orders of both Tand Q.

3. Return to step 1.

Clearly this process will create a mapping that is one-to-one because eachelement is assigned only once. It will also be onto because every elementmust be assigned within a finite number of steps. Namely, ti will certainly beassigned within i steps (or it may have been assigned earlier in the process).This type of argument is called a back-and-forth argument, and this is notthe last time that it will be used.

Also observe that g preserves the ordering of T , and so it follows thatT ' Q.

In the rest of this paper, our goal will to be to generalize the followingtwo results to higher cardinalities. We would like to construct minimalℵα-universal sets for every infinite cardinal ℵα. As we will see, this is almostalways possible.

3 Cofinality

However, before we can construct larger ℵα-universal sets, we first need toreview some more elementary concepts in set theory, beginning with cofinalityand coinitiality.

These two concepts are duals of each other, so we will mostly focus oncofinality. Keep in mind that everything we prove about cofinal sets is alsotrue for coinitial sets, only in reverse.

The notion of cofinality seeks to understand the structure of a totallyordered set. It essentially asks the question, how quickly is a set growingtoward infinity, and how many element must be included to bound thisgrowth [1]?

Definition 15 (Cofinality). Let T be a totally ordered set.

1. Let F ⊆ T . We say that F is cofinal in T if, for every t ∈ T , thereexists at f ∈ F such that t ≤ f .

2. Let ν be an ordinal. Then ν is cofinal in T if S has a cofinalsubset which is isomorphic to ν (i.e. the cofinal subset must be

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well-ordered).

3. Let µ be the least ordinal that is cofinal in T . Then we say that Thas cofinality µ, and we write cf(T ) = µ.

Note that clearly T is cofinal in T , and also that, if T has a greatestelement g, then the set {g} is cofinal in T . We might wonder if it is alwaysthe case that there will be at least one ordinal that is cofinal in T . If thatwas not the case, then the third piece of our definition would not make sense,because we cannot guarantee that such an ordinal exists. However, we neednot worry about this, and the next therom shows [1].

Theorem 16. Let T be a totally ordered set with order ≤. Then thereexists a well-ordered subset W ⊆ T that is cofinal in T , and that satisfies||W || ≤ |T |.

Proof. Recall that there is some well-ordering � of S with the property thatS has ordinality |S| under the order �. Then we can write S as

S = {s1, s2, . . . , s|S|},

where the elements sν are in ascending order under �. We will use thisenumeration to construct our cofinal subset W , however note that W mustbe well-ordered under ≤, not �. To solve this problem, we choose a subsetof S where both orders agree. We define our cofinal subset W as

W = {sν ∈ S | sκ < sν for all κ < ν}.

The set W is ordered under �, and it is a subset of S, and thus it is clearlywell-ordered. Further, observe that ||W || ≤ |S|.

Finally, we claim that W is cofinal in S. To see that this is true, supposeto the contrary that sν is the least element of S that satisfies {sν} > W .Then sκ < sν for all κ < ν, and thus sν ∈W , a contradiction.

Consider the cofinality of a ordinal number ν. From Theorem 16, weknow that cf(ν) ≤ ν. And we can clearly see ν is cofinal in ν. Then aninteresting question is whether there is some other ordinal µ < ν with theproperty that µ is also cofinal in ν.

Definition 17. We say that an ordinal ν is regular if cf(ν) = ν, and wesay that ν is singular if cf(ν) < ν. We also set cf(∅) = 0.

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The ordinals 0 and 1 are regular, however the natural numbers greaterthan one are singular because they have a greatest element, and thus theircofinality is one. The first infinite ordinal (and the first infinite cardinal)ω0 is a regular ordinal because it has cofinality ω0, and the ordinals w0 + 1,w0 + 2, . . . are all singular because they have a greatest element.

It appears that the only zero, one, and infinite cardinal numbers can beregular ordinals because any other ordinal will have a greatest element. Doesthis implication go both ways? That is, are all infinite cardinals regular? Wecan see easily that this is not the case by considering cf(ωω). We know thatthe set S = {ω0, ω1, . . . } is cofinal in ωω, and also that S is order isomorphicto ω0. But this means that cf(ωω) ≤ ω0, and thus ωω cannot be regular.This leads us to the following theorem.

Theorem 18. A regular ordinal is 0, 1, or an infinite cardinal number ℵν .

Proof. We have already seen that a regular ordinal ordinal must be either0, 1, or infinite. Our claim is that all infinite regular ordinals are actuallycardinal numbers.

To that end, suppose µ is an infinite regular ordinal with |µ| = ℵα = να.It follows that να ≤ µ < να+1, and by Theorem 16, there is a well-orderedsubset W ⊆ µ that is cofinal in µ and has cardinality |W | ≤ ωα.

We can then conclude that

cf(µ) ≤ ||W || ≤ ωα ≤ µ, (5)

and since µ is regular, we have that cf(µ) = µ. It then follows that µ =ωα.

We can use this result to show that the cofinality of any totally orderedset must be either zero, one, or a regular initial number.

Let’s recap what we have learned about cofinality: every set S has awell-ordered subset W ⊆ S that is cofinal in S. This cofinal set W itself isregular, and thus it is 0, 1, or an infinite cardinal number ℵν [1].

4 ℵα-universal sets

Now that we have reviewed cofinality, we are in a position to constructℵα-universal sets. Recall that a set S is ℵα-universal if every totally orderedset of cardinality ℵα can be embedded into S.

In this section, we will study a class of sets called ηα-sets. These sets are,in some sense, a generalization of the rational numbers to larger cardinalities.

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In particular, these sets preserve the features that make the rational numbersminimally ℵ0-universal. We will show that, if a few conditions hold, aminimal ηα-set in also minimally ℵα-universal [1].

4.1 ηα-sets

Definition 19. We say that S is an ηα-set if it has no neighboringsubsets which both have a cardinality less than ℵα. The set S is aminimal ηα-set if |S| = ℵα.

This definition sounds quite complicated, but in reality, it just describeshow “densely packed” the set S is. For example, if A and B are neighboringsubsets in an η0-set, then either A or B has cardinality ℵ0. If this is not thecase (i.e. both A and B are finite), then there is some element between Aand B.

For example, Q and R both have this property, and thus they are η0-sets.If the sets A and B were both finite, then A would have a greatest elementa, and B would have a least element b, and so there would be an elementbetween A and B, namely (a+ b)/2.

Suppose that we have successfully constructed an ηα-set. Let us firstverify that it is indeed ℵα universal.

Theorem 20. An ηα-set is ℵα-universal.

Proof. Suppose S is an ηα-set, and A is a totally ordered set with |A| ≤ ℵα.We need to show that A � S.

To do this, we will construct an order preserving injection f : A → Susing transfinite induction. We begin by assigning A a well-order. We canthen use this well-order to list the elements of A:

A = {a0, a1, a2, . . . , a|A|}

We assign f(a0) to any element in S, and now suppose that f has beendefined in such a way that it preserves the order of the first β elements of A.We would like to define f(aβ+1) so that f is still order preserving.

With this goal in mind, we construct two subsets of A given by

A+β+1 = {aν | ν ≤ β and aν > aβ+1},

A−β+1 = {aν | ν ≤ β and aν < aβ+1}.

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We know that |A−β+1| ≤ ℵα and that |A+β+1| ≤ ℵα, and also that

f(A−β+1) < f(A+β+1). It then follows that there is an element s ∈ S with the

property thatf(A−β+1) < s < f(A+

β+1).

Thus we can make the assignment f(aβ+1) = s which maintains the orderpreserving nature of the mapping. Clearly every element in S is assignedonly once, so f is injective, and thus the function f embeds A into S.

We have seen that ηα-sets are ℵα-universal. It clearly follows that aminimal ηα-set will also be minimally ℵα-universal.

Corollary 21. A minimal ηα-set is minimally ℵα-universal.

Now we consider the uniqueness of minimal ηα-sets.

Theorem 22. If A and B are both minimal ηα-sets, then A ' B.

Proof. Let aν and bµ be well-orders of A and B respectively. We will constructan order preserving bijection f : A↔ B using a transfinite back-and-forthargument.

Suppose that some of the elements of A and B have been assigned imagesunder f in such a way that the order of these elements is preserved under themapping. We choose the first element aβ ∈ A that has not yet been assignedunder f , and construct two sets:

A+β = {aν | aν has an imange under f and aµ > aβ}

A−β = {aν | aν has an imange under f and aµ < aβ}.

Notice that A−β and A+β have cardinality less than ℵα because they are both

subsets of A. Also observe that f(A−β ) < f(A+β ) in the set B. Now, because

B is an ηα-set, we can choose b ∈ B such that f(A−β ) < b < f(A+β ). We

make the assignment f(aν) = b which continues to preserve the orders of Aand B.

Now we repeat this process for the first element bβ ∈ B that has notbeen assigned an image, and so on. The mapping f will continue to be orderpreserving, and it will be injective because each assignment uses elements thathave not already been paired. Finally, f is surjective because each elementis guaranteed to be paired after some (transfinite) number of iterations.

Therefore A ' B, and thus minimal ηα-sets are unique.

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To summarize, we know that ηα-sets are ℵα-universal, and that minimalηα-sets are unique.

Another reasonable question we might ask is are minimal ℵα-universalsets unique? The answer is no. For example, both Q, and Q+ ∪ {0}, the setof non-negative rational numbers are minimally ℵ0-universal, however thetwo sets are clearly not order-isomorphic because one has a least elementand the other does not.

4.2 Constructing a minimal ηα-set

Now we turn our attention to the construction of minimal ηα-sets. We willsee that such a set can be constructed whenever ℵα is regular, and a relaxedstatement of the GCH holds.

Consider the set E = {−1, 0, 1} with the standard ordering. Now let Qαbe the collection of all functions f : ℵα → E with the property that for someµ < ℵα, we have f(ν) = 0 for all µ ≤ ν < ℵα. That is, Qα is the set of allfunctions from ℵα to E that are eventually always zero.

We can also view the elements of Qα as tuples of length ℵα. Each entryof each tuple can take on one of the three values in E, and each tuple musteventually become all zeros.

Suppose that f, g ∈ Qα, and that the first element ℵα for which f and gdisagree is is ν. Then we will say that f < g if and only if f(ν) < g(ν).

Theorem 23. Suppose ℵα is a regular cardinal. Then the set Qα is anηα-set.

Proof. Suppose A and B with A < B are neighboring subsets in Qα. Further,suppose that cf(A) < ℵα. Then we need to show that B has a coinitial subsetof order type ℵα. Later, we will argue that the definition of Qα symmetric,and thus all of the arguments we make to construct coinitial subsets in Bcan also be flipped to construct cofinal subsets in A.

We consider three cases: cf(A) = 0, cf(A) = 1, and cf(a) = κ for someregular infinite ordinal κ. First suppose that A = ∅. Then, for each β < ℵα,we define hβ as

hβ(ν) =

{−1 if ν < β,

0 if ν ≥ β.

Clearly the set H = {hβ}β∈ℵα is well-ordered, and has cardinality ℵα. Weknow that ℵα is regular, so the cofinality of B is ℵα, as we had hoped. Allthat remains is to show that H is coinitial in B.

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Suppose f ∈ B is some function, and let ν be the first ordinal for whichf(ν) 6= −1 (we know such a ν exists because f must eventually be alwayszero). Then hν < f , as required.

Now suppose that the set A has a greatest element g (i.e. cf(A) = 1).Let µ be the ordinal for which g(ν) = 0 for all µ ≤ ν ≤ ℵα. Notice that thefirst µ entries of each element in B must be equal to g.

Then for each β < ℵα, we define a function

hβ(ν) =

g(ν) if ν < µ

−1 if κ ≤ ν < µ+ β

0 if µ+ β ≤ ν.

That is, the function hβ is equivalent to g while g is nonzero, followed by βcopies of −1, and then finally all zeros. Once again, the set H = {hβ}β∈ℵαhas cardinality ℵα, so if we can conclude that H is coinitial in B, we will bedone.

Suppose f ∈ B is some function. Let γ be the first entry of f after µthat is not equal to −1. Then hγ < f , as required.

Finally, suppose that cf(A) = κ < ℵα. It follows that A has a cofinalsubset {fδ}δ∈κ where fδ < fε if and only if δ < ε.

Our goal is to construct the supremum of the set A. If we can do this,then the same family of functions that we used in the previous case will be acofinal in B.

To this end, we would like to define two strictly increasing sequences forall ν < κ that are bounded by κ and ℵα respectively. We will denote thefirst sequence as δν < κ, and the second sequence as βν < ℵα. Additionally,we require that these sequences satisfy, for all ν < κ,

1. fδν (γ) = fε(γ) for all ε > δν and for all γ < βν and

2. fδν (βν) < fε(βν) for some ε > δ.

We construct these sequences using transfinite induction. Suppose wehave constructed such a pair of sequences for all ν < µ. We need to constructδµ and βµ.

Let β = sup{βν}ν<µ, and let δ = sup{δν}ν<µ. We choose δµ so thatδ ≤ δµ < κ, and

fδµ(β) = max{fε(β) : δ ≤ ε < κ} ∈ E.

We also choose β ≤ βµ < ℵα such that

βµ = min{γ ≥ β : fδµ(γ) < fε(γ) for some δµ < ε < ℵα}.

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Now we must verify that δµ and βµ actually satisfy our two conditions.However, this is easy because, by construction, βµ is the least ordinal forwhich fδµ is different than a function that is greater than it, and thus bothconditions are satisfied.

Now we will actually find the supremum of the set A. Let θ = sup{βν :ν < κ}. This means that θ is a limit ordinal, and, since ℵα is regular, wehave that θ < ℵα.

We define the function g : θ → E as follows: for each γ ∈ θ, choose aν < κ such that γ < βν , and let g(γ) = fδν (γ). Because of the constructionof our two sequences, this does not depend on which ν we choose.

We then define the same family of functions as in the previous case. Inparticular, we have the functions hβ such that

hβ(ν) =

g(ν) if ν < θ

−1 if κ ≤ ν < θ + β

0 if θ + β ≤ ν.

Clearly this set has cardinality ℵα, so we need to show that it is coinitial inB.

Notice that every element in B, when restricted to θ, must be greaterthan g, and so for the same reason as the previous case, the family hβ willbe coinitial in B.

We have shown that, if cf(A) < ℵα, then we must have coin(B) ≥ ℵα.Now we need to show that the reverse is also true. However, notice thatthe function m : E → E given by m(e) = −e induces an order-reversingbijection on Qα, and thus all of the work we have just done can just as easilybe applied to this new case.

Therefore, Qα is indeed an ηα-set.

We have just shown that Qα is an ηα-set, however, our goal in this sectionwas to construct a minimal ηα-set. We claim that, provided a slightly lessrestrictive version of the GHC holds, Qα is the minimal ηα set.

Theorem 24. Suppose ℵα is a regular ordinal, and that for every ordinalκ < ℵα, we have 2κ ≤ ℵα. Then the totally ordered set Qα is the minimalηα-set.

Proof. We have already seen that, provided that ℵα is a regular ordinal, Qαis an ηα-set. All that remains is to verify that |Qα| = ℵα. Let Qα[ν] be theset of functions in Qα which are always zero after ν. That is

Qα[ν] = {f ∈ Qα | f(µ) = 0 for all ν < µ}.

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Observe that |Qα[ν]| = 3ν ≤ ℵα by our hypothesis, and also observe that

Qα =⋃ν∈ℵα

Qα[ν].

It follows that Qα is the union of ℵα sets, each with cardinality less thanor equal to ℵα, and thus |Qα| = ℵα, as required.

An obvious corollary follows from the fact that ℵ1 is a regular ordinal.

Corollary 25. Suppose the continuum hypothesis holds. Then there is aminimal η1-set.

We have shown that we can construct minimal ℵα-universal sets in somecircumstances. Now we consider the slightly broader question: can we alwaysconstruct a (not necessarily minimal) ℵα-universal set?

The answer is yes. To see why, notice that an ηβ-set is also an ηα-setfor all β ≤ α. This means that, as long as the GHC holds, we can alwaysconstruct an ηβ-set that is also an ηα-set, as we desired.

5 Conclusion

In this paper, we have considered some of the foundations of set theory. Inparticular, we have focused on developing a sufficient amount of mathematicalmachinery so that we could explore some interesting results about universallyordered sets. We proved that minimal ℵα-universal sets exist provided ℵαis regular, and the GCH holds. We also showed that minimal ηα-sets areunique up to order-isomorphism, and that it is always possible to constructan ℵα-universal set, as long as the GCH holds.

This is an extremely rich subfield of set theory, and there are manydirections in which this work could be taken in the future. For example,the construction of ηα-sets is related to the surreal numbers, and alternativeconstruction of the real numbers. Additionally, there are still many interestingquestions remaining about minimal ℵα-universal sets; unlike minimal ηα-sets,these sets are not unique, however, we might conjecture that all minimal ℵα-universal sets share some common core structure. As always in mathematics,more thought is needed.

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References

[1] Egbert Harzheim. Ordered Sets. Springer, 2005. isbn: 978-0-387-24219-4.doi: 10.1007/b104891.

[2] Felix Hausdorff. Grundzuge der Mengenlehre. Leipzig Viet, 1914.

[3] Patrick Keef. An Introduction to the Theory of Sets. 2017.

[4] Patrick Keef and David Guichard. Introduction to Higher Mathematics.2018.

[5] J J O’Connor and E F Robertson. A history of set theory. url: http://www-history.mcs.st-andrews.ac.uk/HistTopics/Beginnings_

of_set_theory.html.

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